the population of a southern city follows the exponential law. If the population doubled in size over an 18 month period and the current population in 10000, what will be the population 2 years from now?
the equation for the exponential equation is: A=ao e ^ -kt
ao is the original amount
a is what the population will be
t is the amount of time
k is the rate of growth
I don't see why the exponent is negative since this an exponentional growth, the negative usually indicates decay.
Anyway...
I would use A = a (e^kt), where t is months
using << the population doubled in size over an 18 month period >>
2 = 1(e^18k)
ln 2 = 18k
k = ln 2/18
so A = 1000 (e^(24*ln2/18))
= 1000(2.51984)
= 2520
another way would be to use
A = 1000(2)t/k and put in t=24
A = 1000(2^1.3333333)
= 2520
Well, let's crunch some numbers, but don't worry, I won't crunch too hard! If the population doubled over an 18 month period, we can say that the original population (ao) was 10,000 and the final population (a) was 20,000. Now, we want to find out what the population will be 2 years from now, which is 24 months.
Using the exponential equation A = ao * e^(-kt), we can rearrange it to solve for k:
a = ao * e^(-k * 18)
20,000 = 10,000 * e^(-k * 18)
Now, let's plug in the values and find k:
2 = e^(-k * 18)
By taking the natural logarithm (ln) of both sides, we can isolate k:
ln(2) = -k * 18
Dividing both sides by -18:
k = ln(2) / -18
Now that we've determined k, we can calculate the population 2 years from now (24 months):
a = ao * e^(-kt)
a = 10,000 * e^(-k * 24)
Plug in the value of k we found earlier:
a = 10,000 * e^((ln(2) / -18) * 24)
Now, let me calculate that for you... *beep boop beep*
The population 2 years from now is approximately 11,296.
So, in a couple of years, our little southern city will be growing, but not too fast!
In this case, we have the following information:
ao (original amount) = 10000 (current population)
t (amount of time) = 2 years
To find the population a (what the population will be) in 2 years, we need to find the rate of growth k.
Given that the population doubled in size over an 18-month period, we can use this information to find k.
Doubling the population means a/ao = 2, or a = 2 * ao.
Since the population doubled in 18 months, we can convert this to years by dividing by 12: t = 18 / 12 = 1.5 years.
Now we can plug these values into the exponential equation and solve for k:
2 * ao = ao * e^(-k * 1.5)
Cancelling out ao on both sides gives:
2 = e^(-k * 1.5)
To solve for k, we can take the natural logarithm (ln) of both sides:
ln(2) = -k * 1.5
Now we can solve for k:
k = -ln(2) / 1.5
Using a calculator, we find k ≈ -0.23105.
Now we have the value of k, we can proceed to find the population in 2 years using the exponential equation:
a = ao * e^(-kt)
a = 10000 * e^(-(-0.23105) * 2)
Using a calculator or a computer, we find that a ≈ 14808.4.
Therefore, the population 2 years from now is approximately 14808.
To find the population 2 years from now, we can use the given information and the exponential equation A = ao e^(-kt).
Let's substitute the given values into the equation. The current population is 10,000, so ao = 10,000. The time period is 2 years, so t = 2.
Now we need to find the rate of growth, k. We are told that the population doubled in size over an 18-month period.
Since the population doubled in size over 18 months, we can use this information to calculate the growth rate.
Forming an equation:
2 = (A / ao) = e^(-k * 1.5)
(A / ao) = 2
e^(-k * 1.5) = 2
To solve for k, we can take the natural log (ln) on both sides:
ln(e^(-k * 1.5)) = ln(2)
-k * 1.5 = ln(2)
Now we can solve for k:
k = -ln(2) / 1.5
Using this value of k, we can calculate the population 2 years from now:
A = ao e^(-kt)
A = 10,000 * e^(-(-ln(2) / 1.5) * 2)
Simplifying further:
A = 10,000 * e^(2 * ln(2) / 1.5)
Calculating the expression:
A ≈ 10,000 * e^(1.3863)
Using a calculator, we find:
A ≈ 10,000 * 3.9925
Therefore, the population 2 years from now would be approximately 39,925.